Critical oxygen deficit and location (table)

The BOD is taken as the 5-day BOD at 20°C. Rates are base $e$ and are corrected to the mixture temperature with $K_T=K_{20}(1.047)^{T-20}$ and $R_T=R_{20}(1.016)^{T-20}$.

Step 1: Mixture properties

$$\begin{aligned} Q &= 0.18+0.46 = 0.64\ \text{m}^3/\text{s}\\ T_{mix} &= \frac{0.18\times 25+0.46\times 22}{0.64} = 22.84^\circ\text{C}\\ DO_{mix} &= \frac{0.18\times 1.5+0.46\times 8.3}{0.64} = 6.39\ \text{mg/l}\\ BOD_5^{mix} &= \frac{0.18\times 32+0.46\times 2}{0.64} = 10.44\ \text{mg/l}\\ DO_{sat}(22.84^\circ\text{C}) &= 8.855\ \text{mg/l (interpolated between 22 and 23}^\circ\text{C in the table)}\\ D_0 &= 8.855-6.39 = 2.47\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and corrected rate constants

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-e^{-5k'}} = \frac{10.44}{1-e^{-5\times 0.2}} = 16.51\ \text{mg/l}\\ k'_T &= 0.2(1.047)^{2.84} = 0.2279\ \text{/day}\\ r'_T &= 0.4(1.016)^{2.84} = 0.4185\ \text{/day} \end{aligned}$$

Step 3: Critical time, deficit and location

$$\begin{aligned} t_c &= \frac{1}{r'-k'}\ln\left[\frac{r'}{k'}\left\{1-\frac{D_0(r'-k')}{k'L_0}\right\}\right] = 2.488\ \text{days}\\ D_c &= \frac{k'L_0}{r'}e^{-k't_c} = 5.10\ \text{mg/l}\\ x_c &= v\,t_c = 0.16\times 86400\times 2.488/1000 = 34.40\ \text{km} \end{aligned}$$

Answer: critical oxygen deficit $D_c = 5.10$ mg/l, occurring about $34.40$ km downstream of point A (after $2.488$ days). The minimum DO is $8.855-5.10 = 3.75$ mg/l.